JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 10)
Let $$\overrightarrow x ,\overrightarrow y $$ and $$\overrightarrow z $$ be three vectors each of magnitude $$\sqrt 2 $$ and the angle between each pair of them is $${\pi \over 3}$$. If $$\overrightarrow a $$ is a non-zero vector perpendicular to $$\overrightarrow x $$ and $$\overrightarrow y \times \overrightarrow z $$ and $$\overrightarrow b $$ is a non-zero vector perpendicular to $$\overrightarrow y $$ and $$\overrightarrow z \times \overrightarrow x ,$$ then
$$\overrightarrow b = \left( {\overrightarrow b \,.\,\overrightarrow z } \right)\left( {\overrightarrow z - \overrightarrow x } \right)$$
$$\overrightarrow a = \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow y - \overrightarrow z } \right)$$
$$\overrightarrow a \,.\,\overrightarrow b = - \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow b \,.\,\overrightarrow z } \right)$$
$$\overrightarrow a = \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow z - \overrightarrow y } \right)$$
Explanation
$\angle \mathrm{AOB}=\angle \mathrm{BOC}=\angle \mathrm{CCOA}=\frac{\pi}{3}$
According to question, we have
$$ \begin{aligned} \vec{a} & =\lambda\{(\vec{x} \cdot \vec{z}) \vec{y}-(\vec{x} \cdot \vec{y}) \vec{z}\} \\\\ & =\lambda\left\{\left(2 \cos \frac{\pi}{3}\right) \vec{y}-\left(2 \cos \frac{\pi}{3}\right) \vec{z}\right\}=\lambda(\vec{y}-\vec{z}) \end{aligned} $$
Thus,
$$ \begin{aligned} \vec{\alpha} & \times(\vec{\beta} \times \vec{\gamma})=(\vec{\alpha} \cdot \vec{\gamma}) \vec{\beta}-(\vec{\alpha} \cdot \vec{\beta}) \vec{\gamma} \\\\ \vec{b} & =\mu\{(\vec{y} \cdot \vec{x}) \vec{z}-(\vec{y} \cdot \vec{z}) \vec{x}\} \\\\ & =\mu\{\vec{z}-\vec{x}\} \end{aligned} $$
Now, $\vec{a} \cdot \vec{y}=\lambda\{2-1\}$; therefore, $\lambda=\vec{a} \cdot \vec{y}$.
Hence, $ \vec{a}=\vec{a} \cdot \vec{y}(\vec{y}-\vec{z})$ ...........(1)
Similarly, $$ \vec{b}=\vec{b} \cdot \vec{z}(\vec{z}-\vec{x}) $$ ............(2)
Now,
$\begin{aligned} \vec{a} \cdot \vec{b} & =(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})\{\vec{y} \cdot \vec{z}-\vec{y} \cdot \vec{x}-\vec{z} \cdot \vec{z}+\vec{z} \cdot \vec{x}\} \\\\ & =(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})\{1-1-2+1\} \\\\ & =-(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z}) .........(3)\end{aligned}$
Hence, from Eqs. (1), (2) and (3), we can conclude that the correct options are $(\mathrm{A}),(\mathrm{B})$ and $(\mathrm{C})$.
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